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Question:

Mukesh, Suresh and Dinesh travel from Delhi to Mathura to attend Janmasthmi Utsav. They have a bike which can carry only two riders at a time as per traffic rules. Bike can be driven only by Mukesh. Mathura is 300Km from Delhi. All of them can walk at 15Km/Hrs. All of them start their journey from Delhi simultaneously and are required to reach Mathura at the same time. If the speed of bike is 60Km/Hrs then what is the shortest possible time in which all three can reach Mathura at the same time.

Explanation:

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​​​​​​​Let Mukesh take Suresh on his bike till B and leave him there to walk till C(Mathura). In the meanwhile, Dinesh keeps walking to reach D, Mukesh comes back picks Dinesh and then both ride to Mathura.

When Mukesh comes back, let us say he meets Dinesh at E.

Let AB = x, then BC = 300 – x

Since Dinesh walks at 15 kmph and bike’s speed is 60 kmph, we have AD = x/4.

∴ BD = $\frac{3x}{4}$

∴ BE = $\left(\frac{60}{75}\right)$$\left(\frac{3x}{4}\right)$ = $\frac{3x}{5}$

Hence, $\frac{300-x}{15}$ = $\frac{{\displaystyle \frac{3x}{5}}+{\displaystyle \frac{3x}{5}}+300-x}{60}$

∴ 4(300 – x) = 300 + x/5

∴ 900 = 4x + $\frac{x}{5}$

$\frac{4500}{21}$ = x

x = $\frac{1500}{7}$ km

Hence, minimum time

= $\frac{x}{60}$ + $\frac{300-x}{15}$

= $\frac{1500}{60\times 7}$ + 20 - $\frac{1500}{15\times 7}$

= $\frac{25}{7}$ + 20 - $\frac{100}{7}$

= 20 - $\frac{75}{7}$

= $\frac{65}{7}$ = 9$\frac{2}{7}$

Hence, option (b).